Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<string, int> > v, v1;
int main() {
  int s = 0;
  int n, m;
  cin >> n >> m;
  int row[n];
  int col[m];
  int row1[n];
  int col1[m];
  fill(row, row + n, 600);
  fill(col, col + m, 600);
  fill(row1, row1 + n, 600);
  fill(col1, col1 + m, 600);
  int a[n][m];
  int a1[n][m];
  int ans1 = 0;
  int ans2 = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      a1[i][j] = a[i][j];
      s += a[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      row[i] = min(row[i], a[i][j]);
      row1[i] = row[i];
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      col[i] = min(col[i], a[j][i]);
      col1[i] = col[i];
    }
  }
  for (int i = 0; i < n; i++) {
    if (row[i] == 0) continue;
    for (int k = 0; k < row[i]; k++) v.push_back(make_pair("row", (i + 1)));
    for (int j = 0; j < m; j++) {
      a[i][j] -= row[i];
      ans1 += (row[i]);
      col[j] = min(col[j], a[i][j]);
    }
  }
  for (int i = 0; i < m; i++) {
    if (col[i] == 0) continue;
    for (int k = 0; k < col[i]; k++) v.push_back({"col", (i + 1)});
    for (int j = 0; j < n; j++) {
      a[j][i] -= col[i];
      ans1 += (col[i]);
      row[j] = min(row[j], a[j][i]);
    }
  }
  for (int i = 0; i < m; i++) {
    if (col1[i] == 0) continue;
    for (int k = 0; k < col1[i]; k++) v1.push_back({"col", (i + 1)});
    for (int j = 0; j < n; j++) {
      a1[j][i] -= col1[i];
      ans2 += (col1[i]);
      row1[j] = min(row1[j], a1[j][i]);
    }
  }
  for (int i = 0; i < n; i++) {
    if (row1[i] == 0) continue;
    for (int k = 0; k < row1[i]; k++) v1.push_back(make_pair("row", (i + 1)));
    for (int j = 0; j < m; j++) {
      a1[i][j] -= row1[i];
      ans2 += (row1[i]);
      col1[j] = min(col1[j], a1[i][j]);
    }
  }
  if (ans1 == s && ans2 == s) {
    if (v.size() <= v1.size()) {
      cout << v.size() << endl;
      for (int i = 0; i < v.size(); i++)
        cout << v[i].first << " " << v[i].second << endl;
    } else {
      cout << v1.size() << endl;
      for (int i = 0; i < v1.size(); i++)
        cout << v1[i].first << " " << v1[i].second << endl;
    }
  } else if (ans1 == s) {
    cout << v.size() << endl;
    for (int i = 0; i < v.size(); i++)
      cout << v[i].first << " " << v[i].second << endl;
  } else if (ans2 == s) {
    cout << v1.size() << endl;
    for (int i = 0; i < v1.size(); i++)
      cout << v1[i].first << " " << v1[i].second << endl;
  } else
    cout << -1 << endl;
}